If a sample of zircon initially contains no Lead, find an expression for the ratio $$\frac{N_U}{N_{Pb}}$$ as a function of time?

How do I do simplifications from this? Do I just go: $$\frac{N_U}{N_{Pb}} = \frac{N_{0(U)}e^{-t_U/\tau_U}}{N_{0(P)}e^{-t_P/\tau_P}}$$

I believe it was also stated that :

The half lives of all the decays leading to Lead are many orders of magnitude smaller than the half life of the initial alpha decay of U.

so I guess $$e^{-t_P/\tau_P} = 1$$

But what is $$\frac{N_{0(U)}}{N_{0(P)}}$$

• A point of vocabulary here: "radiocarbon dating" involves carbon-14. You are using a different decay dating method involving uranium and lead. Carbon doesn't appear anywhere in your problem. – dmckee --- ex-moderator kitten Mar 4 '15 at 2:43
• sorry. i dont have anything else to call it other than dating. – rebc Mar 4 '15 at 2:46
• "radiometric" dating. You are measuring ("metric") radioactive isotopes ("radio"). – CR Drost Mar 4 '15 at 2:48

You might be interested in the extensive Wikipedia article on U/Pb dating. Your equations are a little too simplistic; what's really happened is: at time $t=0$, we had some amounts $N_{\text{Pb}}(0)$ and $N_{\text{U}}(0)$; but due to the $\text{U}\rightarrow\text{Pb}$ decay mode this has changed to:
$$N_{\text U}(t) = N_{\text U}(0) \exp(-t/\tau_{\text U})$$ $$N_{\text{Pb}}(t) = N_{\text{Pb}}(0) + N_{\text{U}}(0) (1 - \exp(-t / \tau_{\text U}))$$
(assuming no decay of Pb; it gets more complicated if Pb is also decaying). Notice that the uranium which hasn't stayed uranium has become lead., hence the $1 -$ that you are missing.
It initially contains no lead, so $N_{0(Pb)} = 0$. Lead is an eventual byproduct of uranium. The statement "The half lives of all the decays leading to Lead are many orders of magnitude smaller than the half life of the initial alpha decay of U." is there to mean that you can assume for practical purposes that Uranium decays into lead, even though it actually decays into something that decays into lead. You need to find out how much uranium is left divided by how much decayed.
• So if we assume that $N(0)Pb = 0$ , The ratio, as above, = $\inf$ ? – rebc Mar 4 '15 at 2:54